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\begin{document}
\title{16.5 习题}
\author{张志聪}
\maketitle

\section*{16.5.1}

\begin{itemize}
  \item (a)

        设$\epsilon > 0$，由傅里叶定理可知，当$N$足够大时，有
        \begin{align*}
          \left\|f - \sum\limits_{n = -N}^N \hat{f}(n)e_n \right\|_2 \leq \epsilon
        \end{align*}

        我们有
        \begin{align*}
           & \sum\limits_{n = -N}^N \hat{f}(n)e_n                                               \\
           & = \hat{f}(0)e_0 + \sum\limits_{n = 1}^N \hat{f}(n)e_n + \hat{f}(-n)e_{-n}          \\
           & = \int_{[0, 1]} f(x) dx
          + \sum\limits_{n = 1}^N (\int_{[0, 1]}f(x)e^{-2\pi n x} dx) e_n
          + (\int_{[0, 1]}f(x)e^{2\pi n x} dx)e_{-n}                                            \\
           & = \frac{1}{2}a_0
          + \sum\limits_{n = 1}^N e_n\int_{[0, 1]}f(x) (cos(2\pi n x) -i sin(2\pi n x)) dx
          + e_{-n}\int_{[0, 1]}f(x) (cos(2\pi n x) +i sin(2\pi n x)) dx                         \\
           & = \frac{1}{2}a_0                                                                   \\
           & + \sum\limits_{n = 1}^N e_n\int_{[0, 1]}f(x) cos(2\pi n x) -i f(x)sin(2\pi n x) dx
          + e_{-n}\int_{[0, 1]}f(x) cos(2\pi n x) +i f(x)sin(2\pi n x) dx                       \\
           & = \frac{1}{2}a_0
          + \sum\limits_{n = 1}^N (e_n + e_{-n})\int_{[0, 1]}f(x) cos(2\pi n x) dx
          + (-e_n + e_{-n})\int_{[0, 1]}i f(x)sin(2\pi n x) dx                                  \\
           & = \frac{1}{2}a_0
          + \sum\limits_{n = 1}^N (e_n + e_{-n})\int_{[0, 1]}f(x) cos(2\pi n x) dx
          + (-e_n + e_{-n})\int_{[0, 1]}i f(x)sin(2\pi n x) dx                                  \\
           & = \frac{1}{2}a_0
          + \sum\limits_{n = 1}^N 2cos(2\pi nx)\int_{[0, 1]}f(x) cos(2\pi n x) dx
          -2isin(2\pi nx)\int_{[0, 1]}i f(x)sin(2\pi n x) dx                                    \\
           & = \frac{1}{2}a_0
          + \sum\limits_{n = 1}^N 2cos(2\pi nx)\int_{[0, 1]}f(x) cos(2\pi n x) dx
          +2sin(2\pi nx)\int_{[0, 1]}f(x)sin(2\pi n x) dx                                       \\
           & = \frac{1}{2}a_0 + \sum\limits_{n = 1}^N a_n cos(2\pi nx) +b_nsin(2\pi nx)
        \end{align*}

        综上可得，$N$足够大时，
        \begin{align*}
          \left\|f - (\frac{1}{2}a_0 + \sum\limits_{n = 1}^N a_n cos(2\pi nx) +b_nsin(2\pi nx)) \right\|_2 \leq \epsilon
        \end{align*}
        所以，$\frac{1}{2}a_0 + \sum\limits_{n = 1}^N a_n cos(2\pi nx) +b_nsin(2\pi nx)$依$L^2$度量收敛于$f$。

  \item (b)

        部分和
        \begin{align*}
           & \sum \limits_{n = -N}^N |\hat{f}(n)|                                                                    \\
           & = \sum \limits_{n = -N}^N |\int_{[0, 1]} f(x)e^{-2\pi n x} dx|                                          \\
           & = \sum \limits_{n = -N}^N |\int_{[0, 1]} f(x)(cos(2\pi nx) - isin(2\pi nx)) dx|                         \\
           & = \sum \limits_{n = -N}^N |\frac{1}{2} (a_n - ib_n)|                                                    \\
           & = |\frac{1}{2} (a_0 - ib_0)| + \frac{1}{2} \sum \limits_{n = 1}^N |(a_n - ib_n)| + |(a_{-n} - ib_{-n})| \\
           & = |\frac{1}{2} a_0| + \frac{1}{2} \sum \limits_{n = 1}^N (|(a_n - ib_n)| + |(a_{-n} - ib_{-n})|)        \\
           & = |\frac{1}{2} a_0| + \frac{1}{2} \sum \limits_{n = 1}^N (|(a_n - ib_n)| + |(a_{n} + ib_{n})|)          \\
           & \leq |\frac{1}{2} a_0| + \sum \limits_{n = 1}^N |a_n| + |ib_n|
        \end{align*}
        于是，由题设$\sum \limits_{n = 1}^\infty a_n, \sum \limits_{n = 1}^\infty b_n$都是绝对收敛，可知
        $\sum \limits_{n = -\infty}^\infty |\hat{f}(n)|$绝对收敛。
        由定理16.5.3可知，
        $\sum \limits_{n = -\infty}^\infty \hat{f}(n)e_n$绝对收敛于$f$。

        于是对任意$\epsilon > 0$，存在$N_0$，使得只要$N \geq N_0$，有
        \begin{align*}
          \left\|f - \sum \limits_{n = -N}^N \hat{f}(n)e_n \right\|_{\infty} \leq \epsilon
        \end{align*}
        由$(a)$可知，
        \begin{align*}
          \left\|f - (\frac{1}{2}a_0 + \sum\limits_{n = 1}^N a_n cos(2\pi nx) +b_nsin(2\pi nx)) \right\|_{\infty} \leq \epsilon
        \end{align*}

        综上，级数$\frac{1}{2}a_0 + \sum\limits_{n = 1}^\infty a_n cos(2\pi nx) +b_nsin(2\pi nx)$一致收敛于$f$。

\end{itemize}

\section*{16.5.2}

\begin{itemize}
  \item (a)

        令$y = 2x - 1$，于是$x = \frac{y + 1}{2}$，函数$y: [-1, 1] \to [0, 1]$，
        于是由命题11.10.7可知，
        \begin{align*}
          a_n & = 2\int_{[0, 1]} f(x) cos(2\pi nx) dx         \\
              & = 2\int_{[0, 1]} (1 - 2x)^2 cos(2\pi nx) dx   \\
              & = \int_{[-1, 1]} y^2 cos(\pi(y + 1)n) dy      \\
              & = \int_{[-1, 1]} y^2 cos(\pi ny)cos(\pi n) dy \\
              & = (-1)^n\int_{[-1, 1]} y^2 cos(\pi ny) dy
        \end{align*}

        $n = 0$时，$a_0 = (-1)^0\int_{[-1, 1]} y^2 dy = \frac{2}{3}$。

        $n \geq 1$时，
        我们设$u = y^2, du = 2ydy; v = \frac{1}{\pi n}sin(\pi ny), dv = cos(\pi ny)$，
        利用命题11.10.1（分部积分法）可知，
        \begin{align*}
          \int_{[-1, 1]} y^2 cos(\pi ny) dy
           & = \int_{[-1, 1]} u dv                                                                  \\
           & = uv|_{-1}^1 - \int_{[-1, 1]} v du                                                     \\
           & = \frac{y^2}{\pi n}sin(\pi ny)|_{-1}^1 - \int_{[-1, 1]} \frac{1}{\pi n}sin(\pi ny)2ydy \\
           & = \frac{2sin(\pi n)}{\pi n} - \frac{2}{\pi n}\int_{[-1, 1]} ysin(\pi ny) dy            \\
           & = - \frac{2}{\pi n}\int_{[-1, 1]} ysin(\pi ny) dy
        \end{align*}
        （以上的处理，可以使$y^2$降幂）

        接下来，我们计算$\int_{[-1, 1]} ysin(\pi ny) dy$。

        令$u = y, du = dy; v = -\frac{cos(\pi ny)}{n\pi}, dv = sin(\pi ny)dy$，再次利用分部积分法，
        \begin{align*}
          \int_{[-1, 1]} ysin(\pi ny) dy
           & = \int_{[-1, 1]} u dv                                                            \\
           & = uv|_{-1}^1 - \int_{[-1, 1]} v du                                               \\
           & = -\frac{ycos(\pi ny)}{n\pi}|_{-1}^1 + \int_{[-1, 1]} \frac{cos(\pi ny)}{n\pi}dy \\
           & = -\frac{2cos(\pi n)}{\pi n} + \frac{1}{n\pi} \frac{sin(\pi ny)}{n \pi}|_{-1}^1  \\
           & = -\frac{2cos(\pi n)}{\pi n}
        \end{align*}

        综上可得，
        \begin{align*}
          a_n & =(-1)^n\int_{[-1, 1]} y^2 cos(\pi ny) dy            \\
              & = (-1)^n (\frac{2}{\pi n}\frac{2cos(\pi n)}{\pi n}) \\
              & = (-1)^n (\frac{4cos(\pi n)}{\pi^2 n^2})            \\
              & = (-1)^n(-1)^n (\frac{4}{\pi^2 n^2})                \\
              & = \frac{4}{\pi^2 n^2}
        \end{align*}

        类似地，我们有
        \begin{align*}
          b_n & = 2\int_{[0, 1]} f(x)sin(2\pi nx) dx        \\
              & = 2\int_{[0, 1]} (1 - 2x)^2 sin(2\pi nx) dx \\
              & = 0
        \end{align*}

        所以，
        \begin{align*}
           & \frac{1}{2}a_0 + \sum \limits_{n = 1}^\infty a_n cos(2\pi nx) + b_nsin(2\pi nx) \\
           & =\frac{1}{3} + \sum \limits_{n = 1}^\infty \frac{4}{\pi^2 n^2} cos(2\pi nx)
        \end{align*}

        因为
        \begin{align*}
          \sum \limits_{n = 1}^\infty a_n
           & = \sum \limits_{n = 1}^\infty \frac{4}{\pi^2 n^2}          \\
           & = \frac{4}{\pi^2}\sum \limits_{n = 1}^\infty \frac{1}{n^2}
        \end{align*}
        由推论7.3.7可知，$\sum \limits_{n = 1}^\infty \frac{1}{n^2}$收敛，
        进一步可知，$\sum \limits_{n = 1}^\infty a_n$绝对收敛。
        又因为$\sum \limits_{n = 1}^\infty b_n$是绝对收敛的。

        综上，由习题16.5.1可知，
        级数$\frac{1}{3} + \sum \limits_{n = 1}^\infty \frac{4}{\pi^2 n^2} cos(2\pi nx)$一致收敛于$f$。

  \item (b)

        因为级数$\frac{1}{3} + \sum \limits_{n = 1}^\infty \frac{4}{\pi^2 n^2} cos(2\pi nx)$一致收敛于$f$，

        对任意$\epsilon > 0$，存在$N_0 > 1$使得只要$N \geq N_0$和$x = 0$，都有
        \begin{align*}
          |(\frac{1}{3} + \sum \limits_{n = 1}^N \frac{4}{\pi^2 n^2} cos(0)) - f(0)| \leq \epsilon \\
          |-\frac{2}{3} + \sum \limits_{n = 1}^N \frac{4}{\pi^2 n^2}| \leq \epsilon                \\
          \frac{2}{3} - \epsilon \leq |\sum \limits_{n = 1}^N \frac{4}{\pi^2 n^2}| \leq \frac{2}{3} + \epsilon
        \end{align*}

        由$\epsilon$的任意性可知，$\sum \limits_{n = 1}^\infty \frac{4}{\pi^2 n^2} = \frac{2}{3}$，
        进一步可得，$\sum \limits_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{4} \times \frac{2}{3} = \frac{\pi^2}{6}$。

  \item (c)

        由题设可知，
        \begin{align*}
          \hat{f}(n) & = \int_{[0, 1]} f(x) e_{-n} dx                         \\
                     & = \int_{[0, 1]} f(x) (cos(2\pi nx) - isin(2\pi nx)) dx \\
                     & = \frac{a_n -ib_n}{2}                                  \\
                     & = \frac{2}{\pi^2 n^2}
        \end{align*}
        （省略了积分的计算过程）

        同理可得，
        \begin{align*}
          \hat{f}(-n) & = \frac{a_n + ib_n}{2} \\
                      & = \frac{2}{\pi^2 n^2}
        \end{align*}
        \begin{align*}
          \hat{f}(0) & = \frac{a_0}{2} \\
                     & = \frac{1}{3}
        \end{align*}

        由命题16.5.4可知，
        \begin{align*}
          \|f\|_2^2 = \sum \limits_{n = -\infty}^\infty |\hat{f}(n)|^2
        \end{align*}

        综上可得，
        \begin{align*}
           & \sum \limits_{n = 0}^\infty |\hat{f}(n)|^2                                                                     \\
           & = |\hat{f}(0)|^2 + \sum \limits_{n = -\infty}^{-1} |\hat{f}(n)|^2 + \sum \limits_{n = 1}^\infty |\hat{f}(n)|^2 \\
           & = \frac{1}{9} + 2\sum \limits_{n = 1}^\infty |\frac{2}{\pi^2 n^2}|^2                                           \\
           & = \frac{1}{9} + 2\sum \limits_{n = 1}^\infty \frac{4}{\pi^4 n^4}                                               \\
           & = \frac{1}{9} + \frac{8}{\pi^4} \sum \limits_{n = 1}^\infty \frac{1}{n^4}
        \end{align*}

        又因为
        \begin{align*}
          \|f\|_2^2 & = \int{[0, 1]} |f(x)|^2 dx   \\
                    & = \int{[0, 1]} (1 - 2x)^2 dx \\
                    & = \frac{1}{5}
        \end{align*}
        综上可得，
        \begin{align*}
          \frac{1}{5} = \frac{1}{3} + \frac{8}{\pi^4} \sum \limits_{n = 1}^\infty \frac{1}{n^4} \\
          \frac{4}{45} = \frac{8}{\pi^4} \sum \limits_{n = 1}^\infty \frac{1}{n^4}              \\
          \sum \limits_{n = 1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}
        \end{align*}
\end{itemize}

\section*{16.5.3}

这道题不对吧！！！都没定义这个特殊符号是怎么运算的。

$P$是一个三角多项式，所以$P \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$，
且存在一个整数$N \geq 0$和一个复数序列$(c_n)_{n = -N}^N$使得$P = \sum \limits_{n = -N}^N c_ne_n$。

\begin{itemize}
  \item (a)
        \begin{align*}
          f \ast P & = f \ast \sum\limits_{n = -N}^N c_n e_n       \\
                   & = \sum\limits_{n = -N}^N f \ast (c_n e_n)     \\
                   & = \sum\limits_{n = -N}^N c_n (f \ast e_n)     \\
                   & = \sum\limits_{n = -N}^N c_n (\hat{f}(n) e_n) \\
                   & = \sum\limits_{n = -N}^N \hat{f}(n) c_n e_n
        \end{align*}
        （书中P345，有类似说明）

        利用引理16.2.5和推论16.3.6，我们有
        \begin{align*}
          \widehat{f \ast P}(n) & = \langle \sum\limits_{k = -N}^N \hat{f}(k) c_k e_k, e_n \rangle \\
                                & = \langle \hat{f}(n) c_n e_n, e_n \rangle                        \\
                                & = \hat{f}(n) c_n \langle e_n, e_n \rangle                        \\
                                & = \hat{f}(n) c_n
        \end{align*}

        由推论16.3.6和定义16.3.7，我们有
        \begin{align*}
          \hat{f}(n) c_n = \langle f, e_n \rangle \langle P, e_n \rangle \\
          \hat{f}(n) \hat{P}(n) = \langle f, e_n \rangle \langle P, e_n \rangle
        \end{align*}
        综上可得，
        \begin{align*}
          \widehat{f \ast P}(n) = \hat{f}(n) c_n = \hat{f}(n) \hat{P}(n)
        \end{align*}

  \item (b) $\circledast$

        由定理16.5.4（Plancherel定理）可知，$|\hat{f}(n)|$对任意$n$有界，
        函数$f(x)$有界。所以存在$M > 0$使得$|\hat{f}(n)| < M$，$|f(x)| < M$。

        由定理16.4.1(三角多项式的魏尔斯特拉斯逼近定理)可知，存在三角多项式$P$使得
        \begin{align*}
          \left\|g - P\right\|_{\infty} \leq \epsilon
        \end{align*}

        由引理16.2.7(b)，我们有
        \begin{align*}
          \left|\widehat{f \ast g}(n) - \widehat{f \ast P}(n)\right|
           & = \left|\langle f \ast g, e_n \rangle - \langle f \ast P, e_n \rangle\right| \\
           & = \left|\langle f \ast (g - P), e_n \rangle \right|                          \\
           & \leq \|f \ast (g - P)\|_2 \|e_n\|_2                                          \\
           & = \|f \ast (g - P)\|_2
        \end{align*}
        又
        \begin{align*}
          \left| f \ast (g - P)\right|
           & = \left| \int_{[0, 1]}f(y) (g - P)(x - y) dy \right| \\
           & \leq \left| \epsilon \int_{[0, 1]}f(y) dy \right|    \\
           & \leq M \epsilon
        \end{align*}

        于是
        \begin{align*}
          \|f \ast (g - P)\|_2
           & = \left(\int_{[0, 1]} |f \ast (g - P)|^2 dx\right)^{1/2} \\
           & \leq M\epsilon
        \end{align*}

        对于
        \begin{align*}
          \left|\widehat{f \ast P}(n) - \hat{f}(n) \hat{g}(n)\right|
           & = \left|\hat{f}(n)\hat{P}(n) - \hat{f}(n) \hat{g}(n)\right|                \\
           & = \left|\hat{f}(n)(\hat{P}(n) - \hat{g}(n))\right|                         \\
           & = \left|\hat{f}(n)(\langle P, e_n \rangle - \langle f, e_n \rangle)\right| \\
           & = \left|\hat{f}(n)\langle P - f, e_n \rangle\right|                        \\
           & \leq |\hat{f}(n)|\left|\langle P - f, e_n \rangle\right|                   \\
           & \leq |\hat{f}(n)|\left\|P - f\right\|_2\left\|e_n \right\|_2               \\
           & = M\epsilon
        \end{align*}

        综上所述，
        \begin{align*}
          \left|\widehat{f \ast g}(n) - \hat{f}(n) \hat{g}(n)\right|
           & = \left|\widehat{f \ast g}(n) - \widehat{f \ast P}(n) + \widehat{f \ast P}(n) - \hat{f}(n) \hat{g}(n)\right|                 \\
           & \leq \left|\widehat{f \ast g}(n) - \widehat{f \ast P}(n)\right| + \left|\widehat{f \ast P}(n) - \hat{f}(n) \hat{g}(n)\right| \\
           & \leq M \epsilon + M \epsilon
        \end{align*}

        由$\epsilon$的任意性可知，$\widehat{f \ast g}(n) = \hat{f}(n) \hat{g}(n)$。

\end{itemize}

\section*{16.5.4}
 (1)

由题设，
我们只需在说明$f^\prime$是1周期函数，即可说明$f^\prime \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$。

对任意$x_0 \in \mathbb{R}$，我们有
\begin{align*}
  \lim\limits_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = f^\prime(x_0)
\end{align*}
于是，对任意$\epsilon > 0$，存在$\delta > 0$，使得只要$|x - x_0| < \delta$，都有
\begin{align*}
  |\frac{f(x) - f(x_0)}{x - x_0} - f^\prime(x_0)| \leq \epsilon
\end{align*}

令$y = x + 1$，
那么对$|y - (x_0 + 1)| < \delta$，即$|x + 1 - (x_0 + 1)| = |x - x_0| < \delta$，
由$f$是1周期函数，我们有
\begin{align*}
  |\frac{f(y) - f(x_0)}{y - (x_0 + 1)} - f^\prime(x_0)|
   & = |\frac{f(x + 1) - f(x_0)}{x + 1 - (x_0 + 1)} - f^\prime(x_0)| \\
   & = |\frac{f(x) - f(x_0)}{x - x_0} - f^\prime(x_0)|               \\
   & \leq \epsilon
\end{align*}
于是可得
\begin{align*}
  f^\prime(x_0 + 1)
   & = \lim\limits_{y \to (x_0 + 1)} \frac{f(y) - f(x_0 + 1)}{y - (x_0 + 1)} \\
   & = f^\prime(x_0)
\end{align*}
所以，$f^\prime$是1周期函数。

(2)

\begin{itemize}
  \item 方法一
        \begin{align*}
          \hat{f^\prime}(n)
           & = \int_{[0, 1]} f^\prime(x) e^{-2\pi i nx} dx \\
        \end{align*}
        令$u = f(x), v = e^{-2\pi i nx}$于是$du = f^\prime (x)dx, dv = -2\pi i n e^{-2\pi i nx}dx$。

        \begin{align*}
          \hat{f^\prime}(n)
           & = \int_{[0, 1]} f^\prime(x) e^{-2\pi i nx} dx                                  \\
           & = uv|_{0}^1 - \int_{[0, 1]} u dv                                               \\
           & = f(x) e^{-2\pi i nx} |_{0}^1 - \int_{[0, 1]} -2 f(x) \pi i n e^{-2\pi i nx}dx \\
           & = f(1) - f(0) + 2\pi i n \int_{[0, 1]} f(x) e^{-2\pi i nx}dx                   \\
           & = 2\pi i n \hat{f}(n)
        \end{align*}

        方法一的问题在于，本书中没有复数函数的导数的定义。胜在方便

  \item 方法二

        利用分部积分法，
        \begin{align*}
          \hat{f^\prime}(n)
           & = \int_{[0, 1]} f^\prime(x) e^{-2\pi i nx} dx                                                          \\
           & = \int_{[0, 1]} f^\prime(x) (cos(2\pi nx) - isin(2\pi nx)) dx                                          \\
           & = \int_{[0, 1]} f^\prime(x) cos(2\pi nx)dx - i \int_{[0, 1]} f^\prime(x) sin(2\pi nx)dx                \\
           & = \left[f(x)cos(2\pi n x)|_0^1 - \int_{[0, 1]} f(x) \cdot (2\pi n)sin(2\pi n x)dx\right]               \\
           & - i\left[f(x)sin(2\pi n x)|_0^1 - \int_{[0, 1]} f(x) \cdot (2\pi n)cos(2\pi n x)dx\right]              \\
           & = \int_{[0, 1]} f(x) \cdot (2\pi n)sin(2\pi n x)dx + i\int_{[0, 1]} f(x) \cdot (2\pi n)cos(2\pi n x)dx \\
           & = 2\pi n \int_{[0, 1]} f(x) sin(2\pi n x) + if(x)cos(2\pi n x) dx                                      \\
           & = 2\pi n \int_{[0, 1]} f(x)(sin(2\pi n x) + icos(2\pi n x))                                            \\
           & = 2\pi n \int_{[0, 1]} f(x)ie^{-2\pi i nx} dx                                                          \\
           & = 2\pi n i \int_{[0, 1]} f(x)e^{-2\pi i nx} dx                                                         \\
           & = 2\pi n i \hat{f}(n)
        \end{align*}

\end{itemize}

\section*{16.5.5}

todo

由Plancherel定理可知，
\begin{align*}
  \|f + g\|_2^2 = \sum\limits_{n = -\infty}^{\infty} |\widehat{f + g}(n)|^2 \\
  \|f - g\|_2^2 = \sum\limits_{n = -\infty}^{\infty} |\widehat{f - g}(n)|^2 \\
\end{align*}
于是，
\begin{align*}
  \|f + g\|_2^2 - \|f - g\|_2^2
   & = \langle f + g, f + g \rangle - \langle f - g, f - g \rangle \\
   & = 2 \langle f, g \rangle + 2\langle g, f \rangle
\end{align*}

我们有
\begin{align*}
   & \sum\limits_{n = -\infty}^{\infty} |\widehat{f + g}(n)|^2 - \sum\limits_{n = -\infty}^{\infty} |\widehat{f - g}(n)|^2                          \\
   & = \sum\limits_{n = -\infty}^{\infty} (\langle f + g, e_n \rangle)^2 - (\langle f - g, e_n \rangle)^2                                           \\
   & = \sum\limits_{n = -\infty}^{\infty} (\langle f, e_n \rangle + \langle g, e_n \rangle)^2 - (\langle f, e_n \rangle - \langle g, e_n \rangle)^2 \\
   & = \sum\limits_{n = -\infty}^{\infty} 4 \langle f, e_n \rangle \langle g, e_n \rangle                                                           \\
   & = \sum\limits_{n = -\infty}^{\infty} 4 \hat{f}(n)\hat{g}(n)                                                                                    \\
\end{align*}

综上可得，
\begin{align*}
  \langle f, g \rangle + \langle g, f \rangle = \sum\limits_{n \in \mathbb{Z}} 2 \hat{f}(n)\hat{g}(n)
\end{align*}


\section*{16.5.6}

先证明函数$f(Lx) \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$。
由$f(x)$是$L$周期函数，我们有，
\begin{align*}
  f(L(x+1)) = f(Lx + L) = f(Lx)
\end{align*}
于是可得$f(Lx)$是1周期函数。

又由$g(x) := Lx$和$f(x)$都连续可知，$f \circ g(x)$连续。

综上可得，$f(Lx) \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$。

\begin{itemize}
  \item (a)

        令$h(x) = f(Lx)$。
        由傅里叶定理可知，对任意$\epsilon > 0$，当$N$足够大时，都有
        \begin{align*}
          \left\| h - \sum_{n = -N}^{N} \hat{h}(n)e_n \right\|_2 < \epsilon                                 \\
          \left\| f(Lx) - \sum_{n = -N}^{N} (\int_{[0, 1]} f(Lx)e^{-2\pi i nx} dx)e_n \right\|_2 < \epsilon \\
          \left(\int_{[0, 1]} |f(Lx) - \sum_{n = -N}^{N} (\int_{[0, 1]} f(Lx)e^{-2\pi i nx} dx)e_n|^2 dx\right)^{1/2} < \epsilon
        \end{align*}
        令$y = Lx$，$y : [0, L] \to [0, 1]$，我们有
        \begin{align*}
          \left(\int_{[0, 1]} |f(Lx) - \sum_{n = -N}^{N} (\int_{[0, 1]} f(Lx)e^{-2\pi i nx} dx)e_n|^2 dx \right)^{1/2} < \epsilon             \\
          \left(\int_{[0, 1]} |f(Lx) - \sum_{n = -N}^{N} (\frac{1}{L}\int_{[0, L]} f(y)e^{-2\pi i ny/L} dy)e_n|^2 dx \right)^{1/2} < \epsilon \\
          \left(\int_{[0, 1]} |f(Lx) - \sum_{n = -N}^{N} c_n e_n|^2 dx\right)^{1/2} < \epsilon                                                \\
          \left(\int_{[0, 1]} |f(Lx) - \sum_{n = -N}^{N} c_n e^{2\pi n i x}|^2 dx\right)^{1/2} < \epsilon
        \end{align*}
        令$z = Lx$，于是
        \begin{align*}
          \left(\int_{[0, 1]} |f(Lx) - \sum_{n = -N}^{N} c_n e^{2\pi n i x}|^2 dx \right)^{1/2} < \epsilon            \\
          \left(\frac{1}{L}\int_{[0, L]} |f(z) - \sum_{n = -N}^{N} c_n e^{2\pi n i z/L}|^2 dz\right)^{1/2} < \epsilon \\
          \int_{[0, 1]} |f(z) - \sum_{n = -N}^{N} c_n e^{2\pi n i z/L}|^2 < L \epsilon^2
        \end{align*}
        由$\epsilon$的任意性和$L$是定值可知，
        $\lim\limits_{N \to \infty} \int_{[0, 1]} |f(x) - \sum_{n = -N}^{N} c_n e^{2\pi n i x}|^2  = 0$。

  \item (b)

        换言之，我们需要证明
        \begin{align*}
          \lim\limits_{N \to \infty}\left\| f - \sum \limits_{n = -\infty}^\infty c_n e^{2\pi i nx / L}\right\|_\infty = 0
        \end{align*}

        利用命题11.10.7（变量替换公式），令$y = Lx$。
        \begin{align*}
          \widehat{f(Lx)}(n)
           & = \langle f(Lx), e_n \rangle                        \\
           & = \int_{[0, 1]} f(Lx) e^{-2\pi i nx} dx             \\
           & = \frac{1}{L}\int_{[0, L]} f(y) e^{-2\pi i ny/L} dy \\
           & = c_n
        \end{align*}
        由题设$\sum\limits_{n = -\infty}^\infty |c_n|$绝对收敛可知，$\sum\limits_{n = -\infty}^\infty |\widehat{f(Lx)}(n)|$绝对收敛。
        于是由定理16.5.3可知，
        \begin{align*}
          \lim\limits_{N \to \infty}\left\| f(Lx) - \sum \limits_{n = -\infty}^\infty  \widehat{f(Lx)}(n)e^{2\pi i nx}\right\|_\infty = 0
        \end{align*}
        替换$y = Lx$，且$c_n = \widehat{f(Lx)}(n)$（这两个值都是定值，替换不会受到影响），我们有
        \begin{align*}
          \lim\limits_{N \to \infty}\left\| f(y) - \sum \limits_{n = -\infty}^\infty  \widehat{f(Lx)}(n)e^{2\pi i ny/L}\right\|_\infty = 0 \\
          \implies                                                                                                                         \\
          \lim\limits_{N \to \infty}\left\| f(x) - \sum \limits_{n = -\infty}^\infty  c_ne^{2\pi i nx / L}\right\|_\infty = 0
        \end{align*}

  \item (c)

        由于$f(Lx) \in C(\mathbb{R}/\mathbb{Z}; \mathbb{C})$，由Plancherel定理可知，
        \begin{align*}
          \|f(Lx)\|_2^2 & = \sum\limits_{n = -\infty}^\infty |\widehat{f(Lx)}(n)|^2 \\
        \end{align*}

        (b)的证明中，有
        \begin{align*}
          \widehat{f(Lx)}(n) = c_n
        \end{align*}

        利用命题11.10.7（变量替换公式），令$y = Lx$。
        \begin{align*}
          \|f(Lx)\|_2^2
           & = \langle f(Lx), f(Lx) \rangle         \\
           & = \int_{[0, 1]} |f(Lx)|^2 dx           \\
           & = \frac{1}{L}\int_{[0, L]} |f(y)|^2 dy
        \end{align*}

        综上可得，
        \begin{align*}
          \frac{1}{L}\int_{[0, L]} |f(x)|^2 dx = \sum\limits_{n = -\infty}^\infty |c_n|^2
        \end{align*}


\end{itemize}



\end{document}


